An ordinary body cools from ' \(4 \theta\) ' to ' \(3 \theta\) ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is \(\theta\) )

According to Newton's law of cooling,
\(\frac{\theta_1-\theta_2}{\mathrm{t}}=K\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right]\)
where, \(\theta_0=\) temperature of surrounding
\(\begin{aligned}
\therefore \quad & \frac{4 \theta-3 \theta}{t}=K\left[\frac{4 \theta+3 \theta}{2}-\theta\right] \\
& \frac{\theta}{t}=K \times \frac{5 \theta}{2}...(i) \\
& K=\frac{2}{5 t}
\end{aligned}\)
After another t min, let the temperature be x.
\(\begin{array}{ll}
\therefore & \frac{3 \theta-x}{t}=\frac{2}{5 t}\left[\frac{3 \theta+x}{2}-\theta\right] \ldots[\text { using (i) }] \\
\therefore & 3 \theta-x=\frac{3 \theta+x-2 \theta}{5} \\
\therefore & 15 \theta-5 x=x+\theta \\
\therefore & 6 x=14 \theta \\
\therefore & x=\frac{7 \theta}{3}
\end{array}\)